Testing Independence In Multi-Event Problems

Why Independence Needs More Than One Check

In a multi-event problem, it is tempting to test one convenient pair and conclude that all the events are independent. That shortcut can fail: independence is a relationship that must be checked across the event system, and pairwise independence alone may not be enough. A separate but related trap is to treat disjoint events as independent. Disjointness means the events cannot occur together; independence means learning that one event occurred does not change the probability of the other.

The key tool is the product identity for intersections. Apply it to every required pair or subset, not just the combination that looks most convenient. The same approach is aligned with the SOA Exam P (Probability) syllabus.

Definition: Two events $A$ and $B$ are independent if $$ P(A\cap B)=P(A)P(B). $$ A collection of events is pairwise independent if this identity holds for every pair in the collection. A collection $A_1,\ldots,A_n$ is mutually independent if, for every nonempty subset of indices $i_1,\ldots,i_k$, $$ P(A_{i_1}\cap\cdots\cap A_{i_k})=P(A_{i_1})\cdots P(A_{i_k}). $$

For three events $A,B,C$, mutual independence requires the three pairwise identities and the three-event identity:

\begin{aligned} P(A\cap B)&=P(A)P(B),\\ P(A\cap C)&=P(A)P(C),\\ P(B\cap C)&=P(B)P(C),\\ P(A\cap B\cap C)&=P(A)P(B)P(C). \end{aligned}

The individual event probabilities are also part of the subset definition, but do not require a special test: for a one-event subset, the identity is simply $P(A)=P(A)$. For three or more events, checking every pair is not sufficient; the larger intersections must also satisfy their product identities.

A Systematic Test

Start by identifying the events and finding their probabilities and intersections from the stated sample space, model, or conditional information. Then list the product identities required by the question. For three events, this means three pairwise checks and one triple check. A failed identity is enough to rule out mutual independence; passing only the pairwise checks establishes pairwise independence, not necessarily mutual independence.

Find the event probabilities.
Compute $P(A_i)$ for each event under the stated probability model.

Find the relevant intersections.
Use the sample space, event definitions, or given probability information to calculate each required intersection.

Compare both sides.
For every required subset, compare the intersection probability with the product of its individual event probabilities.

State the precise conclusion.
Report pairwise independence if all pairs pass; report mutual independence only if every required subset passes.

Why Complement Events Fit the Product Test

The subset identities also let us calculate probabilities involving an event and its complement. For independent $A$ and $B$, for example, the probability that $A$ occurs and $B$ does not is

\begin{aligned} P(A\cap B^c) &=P(A)-P(A\cap B)\\ &=P(A)-P(A)P(B)\\ &=P(A)\bigl(1-P(B)\bigr)\\ &=P(A)P(B^c). \end{aligned}

For three mutually independent events, one can apply the same idea to each event in turn. For example,

\begin{aligned} P(A\cap B\cap C^c) &=P(A\cap B)-P(A\cap B\cap C)\\ &=P(A)P(B)-P(A)P(B)P(C)\\ &=P(A)P(B)\bigl(1-P(C)\bigr)\\ &=P(A)P(B)P(C^c). \end{aligned}

Thus mutual independence gives product probabilities even when some specified events are complements. Repeated subtraction, or inclusion-exclusion, gives the corresponding identity when several events are complemented. This is useful for calculating outcomes such as “$A$ occurs, $B$ does not, and $C$ occurs” without treating that combination as a new independence assumption.

Worked Example: Pairwise Independent but Not Mutually Independent

A hypothetical mechanism produces one of the four binary triples $000, 011, 101, 110$, each with probability $1/4$. Let $A$ be the event that the first entry is 1, $B$ the event that the second entry is 1, and $C$ the event that the third entry is 1. Determine whether the three events are pairwise independent and whether they are mutually independent. The model is a finite sample space with four equally likely outcomes; no interest rate or mortality assumption is involved.

Method. List the outcomes in each event, calculate each event and intersection probability by counting outcomes, and compare with the product identities.

The event sets are $A=\{101,110\}$, $B=\{011,110\}$, and $C=\{011,101\}$. Each contains two of the four equally likely outcomes, so

P(A)=P(B)=P(C)=\frac{2}{4}=\frac12.

The pairwise intersections are $A\cap B=\{110\}$, $A\cap C=\{101\}$, and $B\cap C=\{011\}$. Each contains one outcome, so

P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac14.

For each pair, the product of the individual probabilities is

P(A)P(B)=P(A)P(C)=P(B)P(C)=\frac12\cdot\frac12=\frac14.

All three pairwise identities hold, so the events are pairwise independent. But $A\cap B\cap C$ contains no outcomes, giving $P(A\cap B\cap C)=0$. The product of the three event probabilities is

P(A)P(B)P(C)=\frac12\cdot\frac12\cdot\frac12=\frac18.

Since $0\ne 1/8$, the triple identity fails. The events are not mutually independent. The check is consistent with the sample space: no listed outcome has all three entries equal to 1.

Common mistake avoided: Concluding that the events are mutually independent because all three pairs pass. Pairwise independence checks only subsets of size two; the triple intersection is a separate required check.

Disjointness Is Not Independence

Events $A$ and $B$ are disjoint if $A\cap B=\varnothing$, which implies $P(A\cap B)=0$. For disjoint events to be independent, the product identity would also require $P(A)P(B)=0$. As probabilities are nonnegative, this can happen only if at least one of the events has probability zero. Therefore two disjoint events that both have positive probability are not independent.

The conditional-probability definition makes the distinction intuitive when $P(A)>0$: independence is equivalent to $P(B\mid A)=P(B)$. If $A$ and $B$ are disjoint, then $P(B\mid A)=0$. When $P(B)>0$, this differs from $P(B)$, so learning that $A$ occurred rules out $B$ rather than leaving its probability unchanged.

Worked Example: Disjoint Events in a Ticket Selection

A hypothetical inspection selects one ticket uniformly at random from 20 tickets numbered 1 through 20. Event $A$ is selecting a number from 1 through 5; event $B$ is selecting a number from 6 through 13. Find their probabilities, decide whether they are independent, and calculate the probability of $A\cup B$. The distribution is uniform over the 20 tickets; there is no interest rate or mortality assumption.

There are five tickets in $A$ and eight in $B$, so

P(A)=\frac{5}{20}=\frac14=0.2500,\qquad P(B)=\frac{8}{20}=\frac25=0.4000.

The number ranges do not overlap, so $A\cap B=\varnothing$ and $P(A\cap B)=0$. The product of the event probabilities is

P(A)P(B)=\frac14\cdot\frac25=\frac{1}{10}=0.1000.

Because $0\ne 0.1000$, the events are not independent. Since they are disjoint, their union has probability

P(A\cup B)=P(A)+P(B)-P(A\cap B) =\frac{5}{20}+\frac{8}{20}-0 =\frac{13}{20}=0.6500.

As a conditional check, $P(B\mid A)=0$, while $P(B)=0.4000$. The values differ, confirming dependence. The union calculation is also consistent with the sample space: 13 of the 20 tickets belong to one of the two ranges.

Common mistake avoided: Calling the events independent because they cannot happen together. Their inability to occur together makes their intersection probability zero, which contradicts the positive product $P(A)P(B)=0.1$.

Checking Mutual Independence with Complement Outcomes

A product identity can also give the probability of a mixed occurrence pattern. The next example checks mutual independence from a complete finite model, then uses the complement calculation to find a probability that is not itself one of the original events.

Worked Example: Three Independent Status Indicators

A hypothetical system records three status indicators, each either 0 or 1. Assume the eight possible triples $000,001,\ldots,111$ are equally likely. Let $A,B,C$ denote, respectively, a 1 in the first, second, and third positions. Test mutual independence and calculate the probability that the first and third indicators are 1 while the second is 0. This is the uniform distribution on eight outcomes; no interest rate or mortality assumption applies.

Each event contains the four outcomes with a 1 in its designated position. Therefore

P(A)=P(B)=P(C)=\frac48=\frac12.

For any pair, both designated positions must be 1. There are two possible outcomes for the remaining position, so each pair intersection has probability $2/8=1/4$. For all three events, only $111$ qualifies, so $P(A\cap B\cap C)=1/8$. The products are

P(A)P(B)=P(A)P(C)=P(B)P(C)=\frac14, \qquad P(A)P(B)P(C)=\frac18.

Every pair identity and the triple identity holds. Hence $A,B,C$ are mutually independent. To calculate the requested mixed pattern, there is exactly one qualifying outcome, $101$, among the eight equally likely outcomes. Thus its probability is $1/8=0.1250$. The complement product gives the same result:

P(A\cap B^c\cap C) =P(A)P(B^c)P(C) =\frac12\cdot\frac12\cdot\frac12 =\frac18=0.1250.

The direct count and the product calculation agree, providing a check on both the event interpretation and the arithmetic.

Exam Checks and Common Traps

When a problem asks whether events are independent, state which definition you are testing. For two events, calculate the intersection and the product. For three events, check each pair and the triple intersection before claiming mutual independence. For more events, the definition requires the product identity for every nonempty subset; a convenient sample of subsets is not enough to establish mutual independence.

Pairwise is not mutual. Passing every pairwise check does not establish the identities for larger intersections.
Disjoint is not independent. Disjoint events with positive probabilities have zero intersection but a positive product.
Do not confuse an intersection with a union. Independence is tested with $P(A\cap B)$, not $P(A\cup B)$.
Check the full event definition. When outcomes are equally likely, count the outcomes in each event and intersection using the same sample space.
Use conditional probability carefully. The test $P(B\mid A)=P(B)$ requires $P(A)>0$; the intersection product identity works without that condition.

Key takeaway: Independence is verified by product identities for intersections. Pairwise independence checks every pair; mutual independence checks every nonempty subset. Disjointness instead forces an intersection probability of zero and, when both events have positive probability, rules out independence.

Check Your Understanding

For each problem, identify the relevant product identity before deciding or calculating.

A hypothetical selection process gives $P(A)=0.4$, $P(B)=0.25$, and $P(A\cap B)=0.10$. Are $A$ and $B$ independent? (Answer: Yes, because $P(A)P(B)=0.4(0.25)=0.10=P(A\cap B)$.)
Events $C$ and $D$ are disjoint, with $P(C)=0.2$ and $P(D)=0.5$. Are they independent? (Answer: No; $P(C\cap D)=0$, while $P(C)P(D)=0.10$.)
Three events have probabilities $0.5, 0.4, 0.25$; their pairwise intersection probabilities are $0.20, 0.125, 0.10$, and their triple intersection probability is $0.05$. Are they mutually independent? (Answer: Yes; the pair products are $0.20, 0.125, 0.10$, and the triple product is $0.5(0.4)(0.25)=0.05$.)
Three events have probabilities $0.3, 0.6, 0.5$; every pairwise intersection equals the corresponding product, but $P(A\cap B\cap C)=0.08$. Are they mutually independent? (Answer: No; the required triple product is $0.3(0.6)(0.5)=0.09$, not $0.08$.)
Events $E$ and $F$ are disjoint and independent, and $P(E)=0.2$. Find $P(F)$. (Answer: $P(F)=0$, since $0=P(E\cap F)=P(E)P(F)=0.2P(F)$.)

What You'll Learn